Question:
The value of the definite integral $\int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx$ is:
The value of the definite integral $\int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx$ is:
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For integrals of the form $\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} \, dx$, the answer is always half the upper limit, which is $\frac{\pi}{4}$.
Updated On: Jun 3, 2026
- $\frac{\pi}{4}$
- $\frac{\pi}{2}$
- $\pi$
- $0$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
We apply the definite integral property: \[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Step 2: Meaning
For the limits $0$ to $\frac{\pi}{2}$, the property simplifies to replacing $x$ with $\frac{\pi}{2} - x$. This converts $\sin x$ into $\cos x$ and vice-versa.
Step 3: Analysis
Let the given integral be $I$: \[ I = \int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \quad \text{--- (Eq. 1)} \] Applying the property: \[ I = \int_0^{\pi/2} \frac{\sin^{3/2} (\frac{\pi}{2} - x)}{\sin^{3/2} (\frac{\pi}{2} - x) + \cos^{3/2} (\frac{\pi}{2} - x)} \, dx \] \[ I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} \, dx \quad \text{--- (Eq. 2)} \] Adding Equation 1 and Equation 2: \[ 2I = \int_0^{\pi/2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \] \[ 2I = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]
Step 4: Conclusion
The value of the definite integral is $\frac{\pi}{4}$.
Final Answer: (A)
We apply the definite integral property: \[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Step 2: Meaning
For the limits $0$ to $\frac{\pi}{2}$, the property simplifies to replacing $x$ with $\frac{\pi}{2} - x$. This converts $\sin x$ into $\cos x$ and vice-versa.
Step 3: Analysis
Let the given integral be $I$: \[ I = \int_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \quad \text{--- (Eq. 1)} \] Applying the property: \[ I = \int_0^{\pi/2} \frac{\sin^{3/2} (\frac{\pi}{2} - x)}{\sin^{3/2} (\frac{\pi}{2} - x) + \cos^{3/2} (\frac{\pi}{2} - x)} \, dx \] \[ I = \int_0^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} \, dx \quad \text{--- (Eq. 2)} \] Adding Equation 1 and Equation 2: \[ 2I = \int_0^{\pi/2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \] \[ 2I = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]
Step 4: Conclusion
The value of the definite integral is $\frac{\pi}{4}$.
Final Answer: (A)
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